Find the maximum and minimum values of y= x-2x - Vidyadip

Question

Find the maximum and minimum values of $\text{y}=\tan \text{x}-2\text{x}$

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Answer

We have, $\text{y}=\tan \text{x}-2\text{x}$
$\therefore\ \text{y}=\sec^{2}\text{x}-2$
$ \text{y}=2\sec^{2}\text{x}\tan \text{x}$
For maximum and minimum value, y' = 0
$\Rightarrow\sec^{2}\text{x}=2$
$\Rightarrow\sec\text{x}=\pm\sqrt{2}$
$\Rightarrow\text{x}=\frac{\pi}{4},\frac{3\pi}{4}$
$\therefore \ \text{y}''\Big(\frac{\pi}{4}\Big)=-4<0$
$\therefore \text{x}=\frac{\pi}{4}$ is point of minima.
$ \text{y}''\Big(\frac{3\pi}{4}\Big)=-4<0$
$\text{x}=\frac{3\pi}{4}$ is point of maxima.
Hence, max value
$=\text{f}\Big(\frac{3\pi}{4}\Big)=-1-\frac{3\pi}{2}$
min value
$=\text{f}\Big(\frac{\pi}{4}\Big)=-1-\frac{\pi}{2}$.

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